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Math Help - Related Rates problems

  1. #1
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    Related Rates problems

    I am greatly struggling with related rates problems. I'll give an example and perhaps someone can walk me through it.

    At noon, ship A is 70 km west of ship B. Ship A is sailing south at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 6:00 PM?

    I don't even know where to begin really, according to my notes I should begin by writing down the given info and what I'm trying to find and drawing a picture of applicable, but after doing all this I still have no idea how to go about finding the answer. I know that da/dt would be 25 and db/dt would be 15, but I don't know how I would use that information at all. I also know that a triangle will be involved, but again I don't know how to use that information to solve this problem.
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    Senior Member polymerase's Avatar
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    Ok so bottom line, based on the diagram, you want to find \frac{dz}{dt} when y=240 - that's your question

    So at 6pm, 6 hrs have passed; therefore, ship A has travelled 150km south (y1) and ship B has travelled 90 km north (y2). y= y1+y2=240km

    x = 70km = CONSTANT

    AT THE INSTANT of 6pm, z would be \sqrt{240^2+70^2}=250 km

    Therefore, to solve this question, you use a modified version of pythagorean's theorem: z^2=x^2+(y_1+y_2)^2
    Note: y_1 and y_2 are variables, NOT constants.

    z^2=x^2+(y_1+y_2)^2 \to 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2(y_1+y_2)\left(\f  rac{dy_1}{dt}+\frac{dy_2}{dt}\right)

    Plug in all your numbers and solve... should get \frac{192}{5} km/h
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    I'm still very very confused. I understood everything up to the final two lines, the equation involving z^2=etc. I have no iea how you came up with that to solve it? I understand that the distance between the ships at 6pm is 250 km but I don't understand how to use that information to find the rate of change between the ships at that time.
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by fattydq View Post
    I'm still very very confused. I understood everything up to the final two lines, the equation involving z^2=etc. I have no iea how you came up with that to solve it? I understand that the distance between the ships at 6pm is 250 km but I don't understand how to use that information to find the rate of change between the ships at that time.
    you need to find a relationship between z, x, and y...there's a right angle triangle... z^2=y^2+x^2 seem fimilar to you?
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    Yes of course, but he already used the pythagorean theorem to find the 250km, I don't see how using the pythagorean theorem could possibly lead to the answer of the fraction he provided
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  6. #6
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    Quote Originally Posted by fattydq View Post
    I am greatly struggling with related rates problems. I'll give an example and perhaps someone can walk me through it.

    At noon, ship A is 70 km west of ship B. Ship A is sailing south at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 6:00 PM?
    let ship A begin at the origin. its position as a function of time is ...

    x = 0

    y = -25t

    ship B starts at (70,0). its position as a function of time is ...

    x = 70

    y = 15t

    using the distance formula, the distance between the two ships at any time t is ...

    D = \sqrt{(70-0)^2 + [15t - (-25t)]^2}

    D = \sqrt{4900 + 1600t^2}

    \frac{dD}{dt} = \frac{1600t}{\sqrt{4900+1600t^2}}

    now ... evaluate \frac{dD}{dt} for t = 6 hrs
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  7. #7
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    Quote Originally Posted by skeeter View Post
    let ship A begin at the origin. its position as a function of time is ...

    x = 0

    y = -25t

    ship B starts at (70,0). its position as a function of time is ...

    x = 70

    y = 15t

    using the distance formula, the distance between the two ships at any time t is ...

    D = \sqrt{(70-0)^2 + [15t - (-25t)]^2}

    D = \sqrt{4900 + 1600t^2}

    \frac{dD}{dt} = \frac{1600t}{\sqrt{4900+1600t^2}}

    now ... evaluate \frac{dD}{dt} for t = 6 hrs
    Again I understand the whole thing until you involve the final step, the fraction.
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  8. #8
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    take the derivative of D w/r to t using the chain rule ...

    D = (4900 + 1600t^2)^{\frac{1}{2}}

    \frac{dD}{dt} = \frac{1}{2}(4900 + 1600t^2)^{-\frac{1}{2}} \cdot 3200t

    \frac{dD}{dt} = 1600t(4900 + 1600t^2)^{-\frac{1}{2}}

    \frac{dD}{dt} = \frac{1600t}{\sqrt{4900 + 1600t^2}}

    clear enough?
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